[Chuyên đề] Lượng giác (ver.2)

[lovelycat_handoi95]

Bài 97:[TEX]2cos3xcosx+\sqrt{3}(1+sin2x)=2\sqrt{3}cos^2(2x+\fr{\pi}{4})[/TEX]

Bài 98: [TEX]cotx=tanx+\frac{2cos4x}{sin2x}[/TEX]

Bài 99:[TEX]2sin{\frac{x}{2}}+(sinx-cosx)^2=1+\sqrt{3}cos2x[/TEX]

Bài 100:[TEX]4sin^2\frac{x}{2}-\sqrt{3}cos2x=3-2cos^2(\frac{\pi}{4}-x)[/TEX]

 

[khunjck]

Bài 100:[TEX]4sin^2\frac{x}{2}-\sqrt{3}cos2x=3-2cos^2(\frac{\pi}{4}-x)[/TEX]

Mình nghĩ phần này hạ bâc rồi ra phương trình cuối là:
[TEX]cos (2x - \frac{\pi}{6}) = cosx[/TEX]

[TEX]\left[\begin{x = \frac{\pi}{6} + k2\pi}\\{x =\frac{\pi}{18} + \frac{k2\pi}{3}} [/TEX]

Bài 99:[TEX]2sin{\frac{x}{2}}+(sinx-cosx)^2=1+\sqrt{3}cos2x[/TEX]

[TEX]\Leftrightarrow 2sin{\frac{x}{2}}+1-sin2x=1+\sqrt{3}cos2x[/TEX]
........
[TEX]\Leftrightarrow sin{\frac{x}{2}} =sin(\frac{\pi}{3}+2x)[/TEX]

[TEX]\Leftrightarrow \left[\begin{x =\frac{-2\pi}{9}+k{\frac{4\pi}{3}}}\\{x =\frac{4\pi}{15}+k{\frac{4\pi}{5}}} [/TEX]

 

[thien_nga_1995]

Bài 98: [TEX]cotx=tanx+\frac{2cos4x}{sin2x}[/TEX]

Đk: Sin2x #0
\Leftrightarrow [TEX]\frac {cosx}{sinx} = \frac{sinx}{cosx} + \frac{2cos4x}{sin2x}[/TEX]
\Leftrightarrow [TEX]2cos^2x = 2sin^2x + 2cos4x[/TEX]
\Leftrightarrow[TEX]cos2x = cos4x[/TEX]
\Leftrightarrow.....................

 

[khunjck]

Bài 97:[TEX]2cos3xcosx+\sqrt{3}(1+sin2x)=2\sqrt{3}cos^2(2x+\fr{\pi}{4})[/TEX]

\Leftrightarrow[TEX]cos4x+cos2x+\sqrt3+\sqrt3.sinx=2\sqrt3\frac{1+cos2.(2x+\pi/4)}{2}[/TEX]
.......
\Leftrightarrow[TEX]\sqrt3.sin4x-cos4x = \sqrt3.sin2x + cos2x[/TEX]
........

 

[ngocthao1995]

Bài 97:[TEX]2cos3xcosx+\sqrt{3}(1+sin2x)=2\sqrt{3}cos^2(2x+\fr{\pi}{4})[/TEX]

PT[TEX] \Leftrightarrow 2cos3xcosx+\sqrt{3}+\sqrt{3}sin2x= \sqrt{3}.(1-sin4x)[/TEX]

[TEX]\Leftrightarrow cos4x+cos2x+\sqrt{3}sin2x+\sqrt{3}sin4x=0[/TEX]

[TEX]\Leftrightarrow cos4x+\sqrt{3}sin4x=-(\sqrt{3}sin2x+cos2x)[/TEX]
.......

 

[ngocthao1995]

Câu 101 [TEX]sin4x-cos4x=1+4\sqrt{2}sin(x-\frac{\pi}{4})[/TEX]

Câu 102 [TEX]2\sqrt{2}(sinx+cosx)cosx=3+cos2x[/TEX]

Câu 103 [TEX]sin^4x+sin^4(x+\frac{\pi}{4})+sin^4(x+\frac{\pi}{4})=\frac{9}{8}[/TEX]

Câu 104 [TEX] cosxcos2xcos4xcos8x=\frac{1}{16}[/TEX]

 

[nhockthongay_girlkute]

Câu 101 [TEX]sin4x-cos4x=1+4\sqrt{2}sin(x-\frac{\pi}{4})[/TEX]

[TEX]PT \Leftrightarrow sin 4x - cos 4x = 1+4(sin x - cos x)[/TEX]
[TEX]\Leftrightarrow 2sin 2xcos2x=1+cos 4x +4(sinx - cos x)[/TEX]
[TEX]\Leftrightarrow 2cos 2x (sin 2x - cos 2x)- 4 (sin x - cos x)=0[/TEX]
[TEX]\Leftrightarrow 2(cos^2x - sin^2x)(cos 2x-sin 2x)-4(cosx - sin x)=0[/TEX]
[TEX]\Leftrightarrow 2(cosx-sinx)[(cos x +sin x)(cos 2x- sin 2x)-2]=0[/TEX]
[TEX]\text{ ta co }:(cosx +sin x)(cos 2x- sin 2x)-2=0[/TEX]
[TEX]\Leftrightarrow 2 cos(x-\frac{\pi}{4})cos(2x+ \frac{\pi}{4})-2=0[/TEX]
[TEX]\Leftrightarrow cos 3x +cos(x+\frac{\pi}{2})=2[/TEX]
[TEX]\Leftrightarrow cos 3x + (-sin x)=2[/TEX]
[TEX]\Leftrightarrow \left{-sin x=1\\{cos 3x =1}[/TEX]
[TEX]\Leftrightarrow \left{cosx=0\\{cosx(4cos^2 x -3)=1}(\text{ vo li })[/TEX]

 

[your_ever]

Câu 102 [TEX]2\sqrt{2}(sinx+cosx)cosx=3+cos2x[/TEX]]


[TEX]2\sqrt{2}(sinx+cosx)cosx=3+cos2x[/TEX]

[TEX]\Leftrightarrow2\sqrt{2}sinxcosx+2\sqrt{2}cos^2x=3+cos2x[/TEX]

[TEX]\Leftrightarrow\sqrt{2}sin2x+\sqrt{2}(1+cos2x)=3+cos2x[/TEX]

[TEX]\Leftrightarrow\sqrt{2}sin2x+(\sqrt{2}-1)cos2x=3-\sqrt{2}[/TEX]

...

 

[dandoh221]

Câu 104 [TEX] cosxcos2xcos4xcos8x=\frac{1}{16}[/TEX]

Xét sinx = 0 [TEX] \Leftrightarrow x = k\pi[/TEX] Không thoả mãn
Xét [TEX] sinx \neq 0 [/TEX]. Nhân 2 vế với sinx
[TEX]\Leftrightarrow sinxcosxcos2xcos4xcos8x = \frac{1}{16}sinx[/TEX]
[TEX]\Leftrightarrow \frac{1}{2}sin2xcos2xcos4xcos8x = \frac{1}{16}sinx[/TEX]
[TEX]\Leftrightarrow \frac{1}{16}sin16x = \frac{1}{16}sinx [/TEX]
[TEX]\Leftrightarrow sin16x = sinx[/TEX]

 

[niemkieuloveahbu]

Câu 103 [TEX]sin^4x+sin^4(x+\frac{\pi}{4})+sin^4(x+\frac{\pi}{4})=\frac{9}{8}[/TEX]

[TEX]PT\Leftrightarrow8sin^4x+8sin^4(x+\frac{\pi}{4})+8sin^4(x-\frac{\pi}{4})=9[/TEX]
[TEX]\Leftrightarrow2(1-cos2x)^2+2(1+sin2x)^2+2(1-sin2x)^2=9[/TEX]
[TEX]\Leftrightarrow2cos^22x+4cos2x-1=0[/TEX]
OKnhé, lấy 1 loại 1, nghiệm ra khá lẻ, arccos đó.

 

[buihuukieu]

[TEX]PT\Leftrightarrow8sin^4x+8sin^4(x+\frac{\pi}{4})+8sin^4(x-\frac{\pi}{4})=9[/TEX]
[TEX]\Leftrightarrow2(1-cos2x)^2+2(1+sin2x)^2+2(1-sin2x)^2=9[/TEX]

Dòng này mình k hiểu bn giải thích rõ hộ mình đí

[TEX]\Leftrightarrow2cos^22x+4cos2x-1=0[/TEX]
OKnhé, lấy 1 loại 1, nghiệm ra khá lẻ, arccos đó.

[TEX](1-cos2x)^2 = (2sin^2x)^2 = 4sin^4x[/TEX]

 

[khunjck]

Câu 103 [TEX]sin^4x+sin^4(x+\frac{\pi}{4})+sin^4(x+\frac{\pi}{4})=\frac{9}{8}[/TEX]

\Leftrightarrow[TEX]8.sin^4x+16.sin^4(x+\pi/4)=9[/TEX]
\Leftrightarrow[TEX]8.\frac{1- cos^2{2x}}{2}+16.\frac{1- cos^2{2.(x+\pi/4})}{2}=9[/TEX]
\Leftrightarrow[TEX]4.(1- cos^2{2x}) + 8.(1- cos^2({2x+\pi/2}))=9[/TEX]
\Leftrightarrow[TEX]3-12.cos^2{2x} + 8.sin^2{2x}[/TEX]=0

 

[lovelycat_handoi95]

Bài 105:[TEX]tan^2x-(\sqrt{3}+cot3x)tanx+\sqrt{3}cot3x=0[/TEX]

Bài 106:[TEX]sin(3x+\frac{5\pi}{2})-cos4x+sin(5x+\frac{\pi}{2})=0[/TEX]

Bài 107 :[TEX]2sin^2x+sin2x=2\sqrt{2}sinxsin(3x+\frac{\pi}{4})[/TEX]

Bài 108 :[TEX]cos2x+cos^2x+(5-3cosx)(sinx+cosx)-2 =0[/TEX]

 

[nuhoangbongdem95]

Bài 108 :[TEX]cos2x+cos^2x+(5-3cosx)(sinx+cosx)-2 =0[/TEX]

\Leftrightarrow[TEX] 2cos^2x-1+cos^2x+5sinx+5c0sx-3sinxcosx-3cos^2x-2=0[/TEX]

\Leftrightarrow[TEX] 5(sinx+cosx)-3(sinxcossx+1)=0[/TEX]

Đặt [TEX]sinx+cosx=t[/TEX] là ngon !
............

 

[trang_1995]

Bài 106:[TEX]sin(3x+\frac{5\pi}{2})-cos4x+sin(5x+\frac{\pi}{2})=0[/TEX]

[TEX]pt \Leftrightarrow sin(\frac{\pi}{2}-(-2\pi-3x))-cos4x+cos5x=0 [/TEX]

[TEX]\Leftrightarrow cos(2\pi+3x)-cos4x+cos5x=0[/TEX]

[TEX]\Leftrightarrow cos3x+cos5x-cos4x=0[/TEX]

[TEX]\Leftrightarrow 2cos4xcosx-cos4x=0[/TEX]

[TEX]\Leftrightarrow cos4x(2cosx-1)=0[/TEX]

[TEX]\Leftrightarrow \left\{ \begin{array}{I} cos4x=0 \\ 2cosx-1=0 \end{array} \right[/TEX]......

 

[your_ever]

Bài 107 :[TEX]2sin^2x+sin2x=2\sqrt{2}sinxsin(3x+\frac{\pi}{4})[/TEX]

[TEX]2sin^2x+sin2x=2\sqrt{2}sinxsin(3x+\frac{\pi}{4})[/TEX]

[TEX]\Leftrightarrow1-cos2x+sin2x=-\sqrt{2}[cos(4x+\frac{\pi}{4})-cos(2x+\frac{\pi}{4})][/TEX]

[TEX]\Leftrightarrow1-cos2x+sin2x=-(cos4x-sin4x-cos2x+sin2x)[/TEX]

[TEX]\Leftrightarrow1+cos4x-sin4x-2cos2x+2sin2x=0[/TEX]

[TEX]\Leftrightarrow2cos^2{2x}-2sin2xcos2x-2cos2x+2sin2x=0[/TEX]

[TEX]\Leftrightarrow(cos2x-sin2x)(cos2x-1)=0[/TEX]

ok

[TEX]105:tan^2x-(\sqrt{3}+cot3x)tanx+\sqrt{3}cot3x=0[/TEX]

[TEX]tan^2x-(\sqrt{3}+cot3x)tanx+\sqrt{3}cot3x=0[/TEX]

[TEX]\Leftrightarrow tan^2x-\sqrt{3}tanx-tanxcot3x+\sqrt{3}cot3x=0[/TEX]

[TEX]\Leftrightarrow tanx(tanx-\sqrt{3})-cot3x(tanx-\sqrt{3})=0[/TEX]

[TEX]\Leftrightarrow(tanx-\sqrt{3})(tanx-cot3x)=0[/TEX]

ok

 

[ngocthao1995]

Bài 109 [TEX]\frac{sin^3{\frac{x}{2}-cos^3{\frac{x}{2}}}}{2+sinx}=\frac{1}{3}cosx[/TEX]

Bài 110 [TEX]sin^22x-cos^28x=sin(\frac{17\pi}{2}+10x)[/TEX]

Bài 111 [TEX]cos7x-sin^22x=cos^22x-cosx[/TEX]

Bài 112 [TEX]5sinx-2=3(1-sinx)tan^2x[/TEX]

 

[niemkieuloveahbu]

Bài 92: [TEX]sin4x+sin3x+cosx=4sinx+2[/TEX]

Bài này có lẽ sai đề ạ,em đã kiểm tra,đề của KHTN một lần thì phải,

Thay cho bài này,có bài khá hay,

Bài 113: (=)),trúng số cảnh sát,=)))

[tex] (sinx-2)(sin^2x-sinx+1)=3\sqrt[3]{3sinx-1}+1[/tex]

 

[niemkieuloveahbu]

Bài 109 [TEX]\frac{sin^3{\frac{x}{2}-cos^3{\frac{x}{2}}}}{2+sinx}=\frac{1}{3}cosx[/TEX]

[TEX]PT \Leftrightarrow (sin{\frac{x}{2}}-cos{\frac{x}{2}})(1+\frac{1}{2}sinx)=-\frac{1}{3}(sin{\frac{x}{2}}-cos{\frac{x}{2})(sin{\frac{x}{2}}+cos{\frac{x}{2})[/TEX]

Có nhân tử chung,PT còn lại đặt [TEX]t= sin{\frac{x}{2}}+cos{\frac{x}{2}[/TEX]
DONE!

Bài 110 [TEX]sin^22x-cos^28x=sin({\frac{17\pi}{2}}+10x)[/TEX]

[TEX]\Leftrightarrow -cos4x-cos16x=cos10x\\ \Leftrightarrow cos10x(1+2cos6x)=0[/TEX]

OK!

Bài 111 [TEX]cos7x-sin^22x=cos^22x-cosx[/TEX]

Bài này quái nhở!

[TEX]PT \Leftrightarrow cos4x.cos3x=\frac{1}{2}[/TEX]

Bài 112 [TEX]5sinx-2=3(1-sinx)tan^2x[/TEX]

Đk: [TEX]cosx \neq 0[/TEX]

Biến đổi PT về dạng:
[TEX]2sin^3x+sin^2x-5sinx+2=0\\ \Leftrightarrow sinx=\frac{1}{2}[/TEX]

Lâu lâu không chém,nhớ,=))!

 

[nhockthongay_girlkute]

Bài 113: (=)),trúng số cảnh sát,=)))

[tex] (sinx-2)(sin^2x-sinx+1)=3\sqrt[3]{3sinx-1}+1[/tex]

[TEX]PT \Leftrightarrow sin^3x-sin^2x+sinx-2sin^2x+2sinx-2=3\sqrt[3]{3sinx-1}+1[/TEX]
[TEX]\Leftrightarrow (sinx-1)^3+3(sinx-1)=3sinx-1+3\sqrt[3]{3sinx-1}[/TEX]
[TEX]\text{ Xet} f(t)=t^3+3t [/TEX] đồng biến
\Rightarrow PT có nghiệm [TEX]\Leftrightarrow (sinx-1)^3=3sinx-1[/TEX]